Explanation

This is an algebraic modelling problem with consecutive integers. Let the numbers be n, n+1, n+2, n+3; their squares sum to n² + (n+1)² + (n+2)² + (n+3)² = 4n² + 12n + 14. Setting this equal to 294 gives 4n² + 12n − 280 = 0, i.e. n² + 3n − 70 = 0, which factors as (n + 10)(n − 7) = 0. Since n is natural, n = 7, and the numbers 7, 8, 9, 10 sum to 34. Hence option (b). The distractor 30 corresponds to the set 6–9 (sum of squares 230, too small), and 38 to 8–11 (sum of squares 366, too large).